The DAV Maths Book Class 4 Solutions and **DAV Class 4 Maths Chapter 11 Worksheet 3 **Solutions of Perimeter offer comprehensive answers to textbook questions.

## DAV Class 4 Maths Ch 11 WS 3 Solutions

Question 1.

Solve the following word problems.

(а) Find the perimeter of a rectangle whose length is 12 cm and breadth is 8 cm.

Answer:

Perimeter of a rectangle = 2 l + 2 b

= 2 × 12 cm + 2 × 8 cm

= 24 cm + 16 cm

= 40 cm

Perimeter = 40 cm

(b) Calculate the perimeter of the square whose sides are given below.

(i) 6 cm

(ii) 11 meters

(iii) 26 cm

Answer:

(i) Perimeter of square = 4 × side = 4 × 6 cm

= 24 cm.

(ii) Perimeter of square = 4 × side = 4 × 11 m

= 44 m.

(iii) Perimeter of square = 4 × side = 4 × 26 cm

= 104 cm.

(c) The length of three sides of a triangle are 6 cm, 8 cm and 10 cm. Find its perimeter.

Answer:

Perimeter of triangle = sum of three sides

= 6 cm + 8 cm + 10 cm

= 24 cm.

(d) A boy runs around a rectangular field, the length of which is 35 metres and the breadth is 20 metres. How much distance will be run?

Answer:

To find the distance, we have to find perimeter of rectangular field.

Length of rectangular field = 35 m

Breadth of rectangular field = 20 m

Perimeter of field = 2 × 35m + 2 × 20m

= 70 m + 40 m = 110 m

Boy will run 110 m.

(e) Sheela wants to fix ribbon along the border of a painting. If the length and breadth of the painting are 36 cm and 15 cm respectively how much ribbon is required by Sheela?

Answer:

Since the length and breadth is different so shape of frame is rectangle so we have to find perimeter of frame.

Length of frame = 36 cm

Breadth of frame = 15 cm

Perimeter of frame = 2 × 36cm + 2 × 15cm

= 72 cm + 30 cm

= 102 cm

102 cm ribbon is required.

(f) Raju has to frame a square photo having one side of length 12 cm. How much material will Raju require for this job?

Answer:

We have to find the perimeter of square photo.

Perimeter of square frame = 4 × side

= 4 × 12 cm

= 48 cm

48 cm material is required.

(g) A boy runs two times around a field, the length of which is 62 m and breadth is 40 m. Find the distance covered by him.

Answer:

We have to find perimeter of field.

Length of field = 62 m

Breadth of field = 40 m

Perimeter of the field = 2(62 + 40)

= 2(102)

= 204 m

He runs 2 times so distance covered = 2 × 204 m

= 408 m.

(h) Tina ran around a rectangular garden of length 12 m and breadth 8 m. Meena ran around a square field of side 11m. Who covered more distance and by how much?

Answer:

In both cases we have to find distance means perimeter.

Distance covered by Tina around a rectangular garden = 2 × 12 m + 2 × 8 m

= 24 m + 16 m

= 40 m

Distance covered by Meena, around a square field = 4 × side

= 4 × 11 m

= 44 m

Hence, Meena covered more distance.

Extra distance = 44 – 40

= 4 m

### DAV Class 4 Maths Chapter 11 Worksheet 3 Notes

Perimeter of Rectangle, Square And Triangle

(i) Rectangle:

Opposite sides of the rectangle are equal so the perimeter will be

= l + b + l + b

= 2l + 2 b

= 2 (l + b) unit

Let l = 3 cm, b = 2 cm,

Perimeter 2(3 + 2)

= 10 cm

(ii) Square:

Let the side of square be a since all the sides are equal.

The perimeter = a + a + a + a.

= 4a unit

Let side = 4 cm

Perimeter = 4 × 4

= 16 cm

(iii) Triangle

Perimeter of the triangle will be sum of three sides.

Let the sides be a, b, and c

Perimeter = a + b + c

Let a = 2cm, b = 3cm and c = 4cm

Perimeter = 2 + 3 + 4

= 9 cm.

Question 1.

Babita’s grandfather has a rectangular garden whose length is 70 m and breadth is 40 metres. How much wire is needed to fence the field?

Answer:

The length of wire needed will be equal to perimeter of field.

Length of field = 70 m

Breadth of field = 40 m

Length of wire needed = 2 × 70m + 2 × 40m

= 140 m + 80 m

= 220 m

220 m wire is needed.

Question 2.

Meena wants to fix a border of a square carpet whose each side is 9 m. Find the length of border.

Answer:

Length of border is equal to perimeter of square.

Each side of carpet = 9 m

Length of border = 4 × 9 m = 36 m

36 m length is needed to border the carpet.